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As an exercise I am calculating some orders of some elements but I don't know if my method is correct.

Please can you tell me if my solutions (not the results but the steps to deduce the results) are correct?

The first exercise is this:

If $|a^5| = 12$, what are the possibilities for $|a|$? If $|a^4| = 12$, what are the possibilities for $|a|$?

My method:

In general, if $|a|=n$ then $|a^k|={n\over \gcd(n,k)}$. Using this we can derive:

$$ 12 = {n\over \gcd(n,5)} \tag{$\ast$}$$

We also know that $a^{12 \cdot 5} = a^{60} = e$ and therefore $n \mid 60$. Divisors of $60$ are $1,2,3,4,5,6,10,12,15,20,30 $ and $60$.

From $(\ast)$ we know that $n$ is a multiple of $12$. This narrows down the possibilities for $n$ to:

$$ n \in \{12, 60\}$$

If $n=60$ then $|a^5| = 12$. Also, if $n=12$ then $|a^5|=|a|=12$. Therefore the possibilities for $n$ are $12$ and $60$.

Now for the equation $|a^4|=12$:

By the same argument as before we deduce

$$ 12 = {n\over \gcd(4,n)}$$

and $a^{4 \cdot 12} = a^{48}=e$ and therefore $n\mid 48$. Divisors of $48$ are $1,2,3,4,6,8,12,16,24$ and $48$. Again we know from the equation above that $n$ is a multiple of $12$. This narrwos down the possibilities to $$ n \in \{12,24,48\}$$

If $n=12$ then $|a^4|=3$ so this is not possible. Similarly, if $n=24$ then $|a^4|=6$. Therefore the only possible $n$ is $48$.

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  • $\begingroup$ Your reasoning is correct. $\endgroup$ – kaiten Dec 31 '15 at 2:21
  • $\begingroup$ @kaiten Thank you for your help! $\endgroup$ – a student Jan 5 '16 at 5:34

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